Question: Evaluate $~~\int x\cos x\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $ x\sin x+\cos x+C$ (Choice B) B $ x\sin x-\cos x+C$ (Choice C) C $ -x\sin x+\cos x+C$ (Choice D) D $ -x\sin x-\cos x+C$
Answer: We will solve this by integrating by parts. We know that $ \int u(x)v\,^\prime(x)dx = u(x)v(x)-\int u\,^\prime(x)v(x)dx\,$. We can rewrite this as $ \int u\ dv = uv-\int v\ du\,$. In this problem we will let $~u = x~$ and $~dv=\cos x \,dx\,$. Then $~du = dx~$ and $~v = \sin x\,$. Integration by parts gives $ \int x\cos x\,dx = x\sin x-\int\sin x\,dx\,$ $ ~~~=x\sin x+\cos x+C\,$.